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I'm working in Mathematica and I'm trying to implement the [Pollard's Rho Algorithm for the Discrete Logartihm Problem][1].

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Try using the Which function:

f[0] = 1;
f[x_] := 
  Which[
    Mod[x, 3] == 1, h*f[x-1], 
    Mod[x, 3] == 2, f[x-1]^2, 
    Mod[x, 3] == 0, g*f[x-1] ]

This will take input x and return the first condition which evaluates true

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You can chain Ifs together: If[(condition1), (something), If[(condition2), (something else), If[(condition3), (another thing), (otherwise)]]] or use Which or Switch but I think you're better off using conditioned patterns (patterns that use /;). For example,

With[{h=foo,g=baz},(* <-- since I don't know the actual values *)
 PollardX[0]=1;
 PollardX[i_]:=h PollardX[i-1]/;Mod[i,3]==1;
 PollardX[i_]:=  PollardX[i-1]^2/;Mod[i,3]==2;
 PollardX[i_]:=g PollardX[i-1]/;Mod[i,3]==0;
 ]

And now

PollardX/@Range[8]
(* yields {foo,foo^2,baz foo^2,baz foo^3,baz^2 foo^6,baz^3 foo^6,baz^3 foo^7,baz^6 foo^14}*)
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  • $\begingroup$ An interesting post about the placement of Condition (/;): 533. $\endgroup$ – Sjoerd C. de Vries Oct 12 '15 at 22:03
  • $\begingroup$ That's a good discussion! I usually read /; test as "...only if test", which nicely (in my mind) goes at the end of sentences. $\endgroup$ – evanb Oct 13 '15 at 1:12

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